In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal. Prove it.

Let ABCD be the cyclic trapezium in which AB || DC, AC and BD are the diagonals.

It's seen that chord AD subtends ∠ABD and chord BC subtends ∠BDC at the circumference of the circle.

But, ∠ABD = ∠BDC [Alternate angles, as AB || DC with BD as the transversal]

∴ Chord AD must be equal to chord BC [As equal chord subtends equal angles at circumference]

From figure,

⇒ ∠B + ∠D = 180° [As, sum of opposite angles in a cyclic quadrilateral = 180°]

Also,

⇒ ∠B + ∠C = 180° [Sum of co-interior angles = 180° (As, AB || CD)]

∴ ∠B + ∠C = ∠B + ∠D

⇒ ∠C = ∠D

Now, in ∆ADC and ∆BCD

⇒ DC = DC [Common]

⇒ ∠C = ∠D [Proved above]

⇒ AD = BC [Proved above]

Hence, by SAS criterion of congruence

∆ADC ≅ ∆BCD

∴ AC = BD [By C.P.C.T.]

Hence, proved above AC = BD and AD = BC.